Linked Questions

18
votes
4answers
983 views

Closed form of $I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx$

Does the integral below have a closed-form: $$I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) - 1 - \sqrt{2}} \bigg) \tan(x)\;dx,$$ where $\tan^{-1} (\cdot)$ is inverse tangent function. ...
16
votes
3answers
739 views

Prove that $\int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\frac{\pi}{9} (2160 - 1247\sqrt{3})$

Prove that $$\int_0^\pi\frac{\cos x \cos 4x}{(2-\cos x)^2}dx=\frac{\pi}{9} (2160 - 1247\sqrt{3})$$ I tried to use Weierstrass substitution but the term $\cos 4x$ made horrible algebraic-forms since $...
21
votes
3answers
484 views

Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$

I have a problem with the following integral: $$ \int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\, {{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,} $$ The first idea was to use the ...
12
votes
4answers
450 views

Finding $\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x$

How do we prove that $$I(m)=\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x=\frac{\log{m}}{m^2-1}$$ I see that $$I(m)=\frac{\partial}{\partial m} \int_{0}^{\pi/2} \arctan({m\tan x}) \ \...
9
votes
4answers
459 views

Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$

Calculate the definite integral $$ I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx $$ given that $a>b>0$. My Attempt: If we replace $x$ by $C$, then $$ I = \int_{0}^{\...
8
votes
3answers
1k views

A difficult integral evaluation problem

How do I compute the integration for $a>0$, $$ \int_0^\pi \frac{x\sin x}{1-2a\cos x+a^2}dx? $$ I want to find a complex function and integrate by the residue theorem.
19
votes
1answer
308 views

Evaluating by real methods $\int_0^{\pi/2} \frac{x^5}{2-\cos^2(x)}\ dx$

$\def\Li{{\rm{Li}}}$I'm sure you guys can briefly get the result by some methods of complex analysis, but now I'm only interested in real analysis methods of proving the result. What would you propose ...
8
votes
1answer
333 views

A closed-form of $\frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx$

I am looking for a closed-form of this integral \begin{equation} \frac{1}{2}\int_0^\infty\left[\frac{x^2\cos x}{\cosh 2x-\cos x}-\frac{2x^2}{e^{4x}-2e^{2x}\cos x+1}\right]\,dx \end{equation} I can ...
9
votes
2answers
154 views

A question on cosine integral

So I've read a book and found myself stumped in this integral: $$\int_{0}^{\pi} \frac{\cos(n\theta)}{b^2-a^2\cos(2\theta)}\, d\theta=\begin{cases} \,\,0 &,\quad\mbox{if}\,\, n\,\,\mbox{is odd}\\...
7
votes
1answer
231 views

The elementary methods to compute $\int_0^\pi\frac{e^{ix}}{x-\alpha e^{ix}}\,dx\quad;\quad\text{for}\, \alpha>0$

How to compute the following integral using elementary methods (high school methods). \begin{equation}\int_0^\pi\frac{e^{ix}}{x-\alpha e^{ix}}\,dx\qquad;\qquad\text{for}\, \alpha>0\end{...